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Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.

A closed fermion loop carries an additional minus sign from the odd Grassmann parity of the fields in the Wick theorem contraction expansion. Equivalently, integrating out Grassmann fields gives a determinant, whose logarithmic interaction expansion has the fermionic sign relative to a bosonic inverse determinant. This sign must be included in addition to propagator and vertex phases.

For $C=B^{-1}$, ordered Grassmann Gaussian contractions give $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$ and $⟨(\bar{\theta }M\theta )(\bar{\theta }N\theta )⟩=\mathrm{tr}(MC)\mathrm{tr}(NC)-\mathrm{tr}(MCNC)$. The disconnected contribution has two fermion loops; the connected contribution has one, giving its minus sign. This covariance is an algebraic normalized Berezin integral, rather than a positivity statement about a probability measure.

A closed chain of Dirac propagators is a fermion loop. Its closed fermion loop sign distinguishes it from a bosonic loop. Tracing spinor indices produces gamma matrix traces, and the high-momentum powers of the propagators determine possible ultraviolet divergences.

For a charged Dirac field excitation of mass $m$, multiplicity $N$, electric coupling $eQ$ and scalar Yukawa interaction $-yh\bar{\psi }\psi$, two orientations of a fermion loop induce a Higgs boson coupling to two photons. With metric $(+---)$ and the usual $S=1+iT$ convention, their physical on-shell scattering amplitude isThe Ward identity holds on both photon legs. At low external energy, the integral tends to $1/(3m^2)$. Thus fixed $y$ gives an inverse-mass coefficient, whereas $y=m/v$ gives a constant coefficient as $m$ increases at fixed $v$. The distinction is essential when applying heavy-field decoupling.

A Feynman diagram with one independent closed cycle and hence one independent loop momentum. An $n$-vertex polygon of Dirac propagators in four dimensions has superficial degree of divergence $4-n$ before symmetry cancellations. For a fermion loop, the closed fermion loop sign is separate from any symmetry factor or pairwise charge conjugation cancellation.

With zero scalar sextic coupling, the only potentially logarithmically divergent one-loop six-scalar vertex has three quartic Yukawa vertices on a fermion loop. Its leading numerator is $(/k{)}^3=k^2/k$, whose trace vanishes. A mass insertion or external momentum lowers the degree and makes the integral convergent. This is not vanishing of every odd gamma-matrix trace: for ${\gamma }^0={\sigma }_3$, ${\gamma }^1=i{\sigma }_1$, ${\gamma }^2=i{\sigma }_2$, one has $\mathrm{tr}({\gamma }^0{\gamma }^1{\gamma }^2)=-2i$. No one-loop sextic counterterm is needed, although a finite six-point vertex may remain.

For an unbroken parity symmetry in quantum field theory with $\varphi$ a pseudoscalar, pure-scalar local counterterms of mass dimension at most four are even in $\varphi$. A four-vertex fermion loop requires a quartic counterterm, even when the initial Lagrangian density shows only a Yukawa interaction. The one-point and three-point pure-boson vertices vanish at all orders: their available external four-momenta cannot form a parity-odd Lorentz scalar.

Use Minkowski metric $g=\mathrm{diag}(1,-1,-1,-1)$ and the convention $S_{fi}={\delta }_{fi}+i(2\pi {)}^4{\delta }^{(4)}(p_f-p_i){\mathcal{M}}_{fi}$. Put $m=m_t$, $y={\lambda }_t/\sqrt{2}$, $Q=e_t=2/3$ andThe Feynman i-epsilon prescription is understood in subsequent denominators. Contracting the spinor indices around the fermion loop gives the traceHere $/p={\gamma }^{\rho }{p}_{\rho }$ is Feynman slash notation. The closed fermion loop sign is $-1$, the three interaction vertices contribute $(-iy)(-ieQ{)}^2$, and the three Dirac propagators contribute $i^3$. Their product, including the color charge trace, is $-N_{c}y{e}^2{Q}^2$. Thus one ordering hasMultiplying by the incoming polarization vectors and adding the exchanged Feynman diagram givesThe displayed four-dimensional Feynman integral is formal until regularized; subsequent manipulations use dimensional regularization, which permits the required loop momentum shifts.

There are two one-loop Feynman diagrams: the two orders of the labelled photons around a top quark fermion loop, with one Yukawa interaction insertion for the outgoing Higgs boson. They are added, rather than identified by a Feynman-diagram symmetry factor. Each has the same color charge trace $N_c=3$.

In the first routing, traverse the fermion loop from the $p_2$ insertion to the $p_1$ insertion with momentum $l$, then to the Higgs boson insertion with momentum $l+p_1$, then back to the $p_2$ insertion with momentum $l-p_2$. Conservation at the Higgs boson insertion gives $p_h=p_1+p_2$. The second routing exchanges $(p_1,\mu )$ and $(p_2,\nu )$: its three internal momenta are $l$, $l+p_2$ and $l-p_1$. Reversing the orientation is another way to represent this second ordering. Two vector insertions give no Furry theorem cancellation.

At second order in the interaction term, attach the two external fermion fields to different vertices and contract the remaining two fermion fields with each other and the two vector fields with each other. The factor $1/2!$ in the expansion is cancelled by the two ways to assign the external ends. There is no closed fermion loop, and hence no extra closed-loop fermionic sign.

After removing the two external Dirac propagators, the internal momentum assignment $l$ for the photon and $p+l$ for the fermion gives the one-loop insertionThus the insertion is proportional to exactly the propagator product specified in the source. Define $\mathcal{I}=-i\Sigma$. With Fourier convention $\psi (x)\sim e^{ip\cdot x}$, the reduced free inverse kernel is $K_0(p)=-/p-m$, since the printed Dirac propagator equals $i{K}_0^{-1}$. Repeated insertions giveThe factors of $i$ and the sign of the named self-energy are convention dependent; the cancelling counterterm is fixed once this insertion convention is specified. The internal photon joins two vertices on one directed fermion line, as in the first panel of the figure in 1b.

The massive Dirac propagator isUse the gamma matrix trace identities $\mathrm{tr}{\gamma }^{\mu }=0$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\rho })=4g^{\mu \rho }$. They giveThe integrand is odd under $k\mapsto -k$, and dimensional regularization preserves that symmetry. The same vanishing holds for $S=i{S}_F$. One must not assert that $S(0)$ itself is zero: its term proportional to $m1$ generally is nonzero and divergent, but its trace against a single gamma matrix is zero. This one-photon fermion loop is also the simplest instance of Furry's theorem.

First fix a convention that is implicit in the source. For the stated free generating functionalthe time-ordered Wick contractions are $S_F=⟨T\psi \bar{\psi }{⟩}_0=-iS$ and ${\Delta }_F=⟨TAA{⟩}_0=-i\Delta$. These follow directly by the source differentiation rules in part (i). Thus the kernels called propagators in the question are distinguished here from the contractions, which already include their factors of $i$.

At first order in the QED interaction, write $a$ for its integration point. The normalized three-point correlation function isThe zeroth-order term vanishes because the free photon one-point correlation function is zero. There are two nonzero Wick contractions of the four fields:The first term pairs the two external fermions directly and closes the two fields at the interaction into a fermion loop; its minus sign is the closed fermion loop sign. The second term joins the external fermions through the interaction, giving the connected contribution. This is also obtained by differentiating the quartic term $(i\bar{\eta }S\eta {)}^2/2$ in the free generating functional; the two choices of bilinear cancel its factor $1/2$.

Insert $S_F=-iS$ and ${\Delta }_F=-i\Delta$. Since $(-ie)(-i{)}^3=e$, the result is exactlyThe trace term is a disconnected product of a photon tadpole diagram with the free external fermion propagator; the other term is the connected three-point correlation function. In this QED calculation the disconnected term vanishes, as established next. With a convention using $S_F,{\Delta }_F$ themselves as the symbols for the propagators, the equivalent formula retains the prefactor $-ie$ and the connected-minus-trace ordering above; mixing these two conventions would give an erroneous overall phase.

There are exactly four tree-level Feynman diagrams: one four-photon contact graph and one exchange graph for each of the three partitions $12|34$, $13|24$ and $14|23$ of the external legs. The exchange graphs contain two cubic Feynman vertices. A fermion loop or a closed Faddeev-Popov ghost line has positive loop order, so neither enters here.

Write $p_1+p_2+p_3+p_4=0$ and take physical photon polarization vectors ${\epsilon }_i$ with $p_i\cdot {\epsilon }_i=0$ and $p_i^2=0$. At the cubic Feynman vertex attached to legs $i,j$, the two terms containing $p_i\cdot {\epsilon }_i$ or $p_j\cdot {\epsilon }_j$ vanish. If $q=p_i+p_j$ flows into the internal line, the remaining factor at that end isAt the other end the internal incoming four-momentum is $q$ rather than $-q$, giving $+2\lambda q^{\nu }({\epsilon }_k\cdot {\epsilon }_l)$. Multiplication by the photon propagator givesThe cancellation holds first at generic nonzero channel Mandelstam invariants and then by continuation at degenerate kinematics. Summing the three exchange graphs produces $4i{\lambda }^{2}P$, whereThe quartic Feynman vertex contributes $-4i{\lambda }^{2}P$. ThereforeThis is a cancellation for physical on shell photon polarization vectors. The off shell four-photon correlation function need not vanish. Physical light-by-light scattering first occurs through charged-particle loop Feynman diagrams.

The two one-loop Feynman diagrams are fermion loops with three photon interaction vertices. Fix vertex 1; the two cyclic orders are $(1,2,3)$ and $(1,3,2)$. A cyclic rotation does not produce another Feynman diagram. Keeping all three external labels in the same positions, these two orders can be represented by the same triangle with opposite fermion flow, as in the first two panels.

The photons are labelled $(k_1,\mu )$, $(k_2,\nu )$, $(k_3,\rho )$, with all four-momenta incoming and $k_1+k_2+k_3=0$. The amputation of external propagators removes the three external photon propagators, leaving their Lorentz indices open. There are no other connected one-loop Feynman diagrams with only three external photons in QED.

For an $n$-vertex fermion loop, reversing the fermion flow reverses the ordered product inside its trace. The same charge-conjugation matrix calculation transforms every Dirac propagator by $p\mapsto -p$ and contributes $-1$ for each vector photon interaction vertex. Relabel the loop momentum so that the external labels are fixed. The reversed Feynman diagram therefore obeysFor odd $n\ge 3$, the $(n-1)!$ cyclic orders pair into opposite orientations and cancel. For $n=1$, the single photon tadpole maps into itself with a minus sign and is zero. The common closed fermion loop sign is present on both sides and does not supply the factor $(-1{)}^n$. This proves Furry's theorem for all the requested one-loop Feynman diagrams. The charge conjugation argument also gives vanishing odd-photon correlation functions to all orders in vacuum QED, provided no charge conjugation symmetry breaking background or charged external state is introduced.

The third panel of the figure in part a is a representative four-photon box diagram. Fix the external photon labelled 1 and orient the fermion loop. The six inequivalent cyclic orders of the labelled interaction vertices areThey are obtained by permuting labels $2,3,4$ on the displayed box, keeping label 1 fixed. Equivalently there are three undirected boxes, each with two opposite fermion flows. The reversal pairs areFor four photon interaction vertices, the sign in part d is positive, so the members of each pair add. They can be represented by three box diagrams multiplied by two. There is no tree photon self-coupling in QED, and therefore no triangle or bubble with a photon self-vertex to add at this loop order. This is the one-loop light-by-light scattering contribution, of order $e^4$.

Let the masses of $\Phi$ and $\varphi$ be $M$ and $\mu$, respectively. Write their external four-momenta as $P=k_1+k_2$, with $P^2=M^2$ and $k_1^2=k_2^2={\mu }^2$. The condition $M>2\mu$ opens the two-body decay phase space. Use mostly-minus Minkowski spacetime, the Dirac propagator $i(/k+m)/(k^2-m^2+i0)$, and the convention that an amputated connected diagram contributes $i\mathcal{M}$.

There is no tree-level three-scalar Feynman vertex in the displayed interaction. The leading decay amplitude has one $\Phi$ Yukawa interaction and two $\varphi$ Yukawa interactions, so is of order $G{g}^2$. The Yukawa fermion triangle is:

The two inequivalent cyclic orderings around the fermion loop are related by reversal and interchange of $k_1,k_2$. They give equal integrals: reversing all internal four-momenta leaves the masses and denominators unchanged, and the scalar gamma matrix trace identities below give the same numerator. Both must be included. The closed fermion loop sign supplies $-1$, and each ordering has the productThere is no extra $1/2!$ in the decay amplitude. Such a factor belongs to integration over the identical-particle final phase space when calculating a decay rate.

Route $a=\ell$, $b=\ell +k_1$, $c=\ell -k_2$ and define $D_a=a^2-m^2+i0$, and similarly for $D_b,D_c$. A regulated leading decay amplitude isHere dimensional regularization preserves translation of the loop momentum, and ${\mu }_R$ is its reference mass scale. Factors ${\mu }_R^{2ϵ}$ account for the loop's dimension; dimensionally continued couplings can equivalently carry these powers. No ${\gamma }^5$ occurs in this scalar loop, so there is no ${\gamma }^5$ regularization ambiguity.

Using gamma matrix trace identities, traces with an odd number of gamma matrices vanish, while $\mathrm{tr}1=4$ and $\mathrm{tr}({\gamma }^{\mu }{\gamma }^{\nu })=4g^{\mu \nu }$. Thus the numerator simplifies completely toFor a reduction to standard scalar integrals, disregard the vanishing infinitesimals only in the numerator. Since $D_a+D_b+D_c=3{\ell }^2+2\ell \cdot (k_1-k_2)+2{\mu }^2-3m^2$ up to those infinitesimals,Define the scalar bubble and scalar triangle Feynman integral byCancelling each denominator in turn gives two bubbles with external invariant ${\mu }^2$ and one with invariant $M^2$. ThereforeThis explicitly displays the complete gamma matrix trace simplification, the two loop orientations and the overall phase convention.

There is a renormalization qualification to this answer. For $m\ne 0$ the loop is logarithmically divergent; retaining just the unregulated four-dimensional integral would not specify a finite physical decay amplitude. To see the divergence, a Feynman parameter and a shifted loop momentum giveThe scalar triangle Feynman integral is ultraviolet finite. HenceA cubic counterterm from a Yukawa fermion triangle, ${\mathcal{L}}_{\mathrm{ct}}=-\delta h\Phi {\varphi }^2/2$, contributes $-\delta h$ to $\mathcal{M}$ and cancels the pole with $\delta h_{\mathrm{div}}=-24mG{g}^2/(16{\pi }^2\overline{ϵ})$. This is an instance of counterterm closure of a massive Yukawa theory. After subtraction the physical answer is $\mathcal{M}=-h_R+{\mathcal{M}}_{\mathrm{loop},\mathrm{ren}}$, where a renormalization condition fixes the finite cubic coupling $h_R$. The displayed interaction alone does not supply that condition. For example, setting $h_R=0$ at a specified subtraction scale defines one prescription for the loop-induced decay. The regulated expressions above remain the lowest-order loop answer without such extra data. For $m=0$ the trace, and therefore this scalar triangle contribution, vanishes identically.

Now take the Hermitian pseudoscalar Yukawa interaction with the sign printed in the PDF. Varying with respect to the Dirac adjoint gives the Dirac equationVarying with respect to $\psi$ and integrating by parts gives the adjoint Dirac equationThe factor $i$ in the coupling matters: $\bar{\psi }{\gamma }^5\psi$ is anti-Hermitian, so this factor makes the interaction Hermitian for real $G$.

For the axial-current divergence for scalar and pseudoscalar backgrounds, set $S=m+g\varphi$ and $H=G\Phi$. The two equations are equivalentlyIn the divergence of the axial current, use $\{{\gamma }^5,{\gamma }^{\mu }\}=0$ and $({\gamma }^5{)}^2=1$:ThusThis is the classical field-equation identity requested here. It includes explicit breaking by both backgrounds; it is not a claim that a renormalized quantum axial current has no chiral anomaly when gauge interactions are present.

For one Grassmann variable, define the Berezin integral by $\int d\theta 1=0$ and $\int d\theta \theta =1$, extending linearly. Since ${\theta }^2=0$, these rules determine it on every polynomial. With several Grassmann variables, integrate from the rightmost differential first. Equivalently, $\int d{\theta }_n\cdots d{\theta }_1$ extracts the coefficient of ${\theta }_1\cdots {\theta }_n$ in that order. Repeated indices give zero, while permuting distinct generators gives the fermionic sign of that permutation. Hencewhere $\epsilon$ is the Levi-Civita symbol.

For the doubled variables, fix the paired measure orientation byThe exponential in the Grassmann Gaussian integral has a finite Taylor expansion. Only its term of degree $2n$ survives integration. Each barred and unbarred index must occur exactly once; the $n!$ orders of the even bilinears cancel the Taylor factorial. Reordering the unbarred indices gives the determinant permutation sign. ThereforeThis identity holds even for a singular matrix. The normalized expectations below require $B$ invertible.

Put $C=B^{-1}$. Differentiating with respect to $B_{ji}$ inserts $-{\bar{\theta }}_j{\theta }_i={\theta }_i{\bar{\theta }}_j$. The derivative of a determinant then gives the ordered Grassmann Gaussian contractionsDifferentiate once more, using $\partial C_{ij}/\partial B_{lk}=-C_{il}{C}_{kj}$. The result isThe minus sign is the interchange sign in the fermionic Wick theorem. In particular $⟨\bar{\theta }M\theta ⟩=-\mathrm{tr}(MC)$. Reordering each barred-unbarred pair in the product and contracting the indices givesThe covariance of Grassmann bilinears is the connected second term. Its Feynman diagram is one oriented fermion loop through both insertions; the first term has two separate fermion loops, one at each insertion. Each closed fermion loop contributes a minus sign, so two loops give a plus and one gives a minus.

For the Dirac field, write $D=\gamma \cdot \partial +m$ and use signature $(-,+,\dots ,+)$, consistent with the denominator in the question. A finite regulator turns the free fermionic path integral into the same Gaussian problem with $B=iD$, because $e^{iS}=e^{-i\bar{\psi }D\psi }$. The time-ordered product covariance is consequently $(i{D}_F{)}^{-1}=-i{D}_F^{-1}$, where the subscript specifies the Feynman i-epsilon prescription.

With the Fourier transform convention $\psi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\tilde{\psi }(p)$, the derivative kernel is $i\gamma \cdot p+m$. The gamma matrices obeysince their anticommutator makes $(\gamma \cdot p{)}^2=p^{2}I$. Thus the mostly-plus Dirac propagator with a real derivative kernel isFor $p^2=-p_0^2+{\mathbf{p}}^2$, the denominator is $p_0^2-{\mathbf{p}}^2-m^2+i0$. Its positive-energy pole is below and negative-energy pole above the real axis, giving time ordering. Here $S_F$ is defined without the leading factor $i$, as in the question; conventions that include that factor in the name Dirac propagator give the same covariance.

Use the printed interaction, so the Dyson series contributes $-ig$ per vertex before its spinor matrix. The momentum-space Feynman rules areEach interaction vertex has one pseudoscalar line and one incoming and one outgoing fermion-number line. Conserve four-momentum at every vertex, integrate each independent loop momentum with $d^4\ell /(2\pi {)}^4$, and divide by any Feynman-diagram symmetry factor. A closed fermion loop contributes an extra minus sign. The external wavefunctions are $u$ for incoming fermions, $\bar{u}$ for outgoing fermions, $\bar{v}$ for incoming antifermions and $v$ for outgoing antifermions; spinor matrices are multiplied in order along the fermion line. An external pseudoscalar has unit wavefunction factor in standard relativistic normalization. Relative fermionic signs also occur when the external contractions are permuted.

The leading scattering amplitudes are of order $g^2$. Label the incoming momenta $p_1,p_2$ and outgoing momenta $p_3,p_4$, all with positive energy and $p_i^2=m^2$. Use the Mandelstam variables $s=(p_1+p_2{)}^2$, $t=(p_1-p_3{)}^2$ and $u=(p_1-p_4{)}^2$. The tree scattering with pseudoscalar exchange is shown below; dashed edges are pseudoscalar propagators, and arrows indicate fermion-number flow, opposite to the motion of an antifermion.

Define the scattering amplitude by $⟨f|S-1|i⟩=i(2\pi {)}^4{\delta }^4(p_3+p_4-p_1-p_2)\mathcal{M}$. Use standard relativistic normalization for external states and spinors. To fix even the overall signs, suppress the positive state-normalization factors and write the creator order of the two-fermion states as $|i⟩=b_1^{†}{b}_2^{†}|0⟩$, $|f⟩=b_3^{†}{b}_4^{†}|0⟩$, and the fermion-antifermion states as $|i⟩=b_1^{†}{d}_2^{†}|0⟩$, $|f⟩=b_3^{†}{d}_4^{†}|0⟩$. Spin labels on the $u_i,v_i$ are implicit.

For two fermions, the direct $t$ graph connects $1\to 3$ and $2\to 4$, while the exchanged $u$ graph connects $1\to 4$ and $2\to 3$. The canonical anticommutation relations give signs $+$ and $-$ for these contractions in the chosen state order. Multiplication of the two vertices and the scalar propagator gives $(-ig{)}^{2}i=-i{g}^2$. HenceInterchanging the labels of either pair of identical external fermions reverses this amplitude, as required for the antisymmetric states of fermionic Fock space.

For a fermion and an antifermion, the graphs are $t$ exchange and $s$ annihilation. Their signs can be derived without guessing an analogy with the first process. The normal-ordered product of the pseudoscalar bilinear contains terms of the formsThe minus in the second term is the antifermion sign of a normal-ordered bilinear. Between the chosen external states, exchange has sign $-$ and pair annihilation followed by pair creation has sign $+$. ThereforeChanging an external-state phase can change the common overall sign; it cannot change the relative minus between the two channels. Both amplitudes use the coefficient exactly as printed. For the Hermitian convention $g=i{g}_P$ discussed in part (a), substitute $g^2=-g_P^2$ throughout, or equivalently use vertex $g_P{\gamma }^5$ from the outset.